Please do Q3 a and c In the laboratory, a student filled a 250-mL container with

Question

Please do Q3 a and c

In the laboratory, a student filled a 250-mL container with an unknown gas un pressure of 760 ton was obtained. The sample of gas was found to weigh 0.164 g. Calculate the molecular mass of the gas if tire temperature in the laboratory was 25 degreeC. The exhaled by an average human being might have the following typical composition, expressed in terms of partial pressure: N_2, 569 torr, O_2, 116 torr, CO_2, 28 tort; water vapor, 47 torr. What are the mole fractions of each gas?

Explanation / Answer

V = 0.25L

P = 1 atm

m = 0.164g

T = 25 +273 = 298K

R = 0.082LatmK-1mol-1

PV = nRT (Ideal gas equation)

n = PV/RT = 1atm x 0.25L/ 0.082LatmK-1mol-1 x 298K = 0.0102

n = m/M (moles = amount/molecular weight)

M = m/n = 0.164/0.0102 = 16g/mol

M = 16g/mol

c) The ratio of the partial pressure to the total pressure is equal to the mole fraction of the component gas.

N2 = 569 torr

O2 = 116 torr

CO2 =28 torr

Water Vapor = 47 torr

Pi/Pt = ni/nt = xi

X(N2) = 569/760 = 0.74

X(O2) = 116/760= 0.15

X(CO2) = 28/760= 0.036

X(wv) = 47/760=0.0618

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